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MAB241 COMPLEX VARIABLES
LAURENT SERIES
1 What is a Laurent series?
The Laurent series is a representation of a complex function f (z) as a series. Unlike the Taylor series which expresses f (z) as a series of terms with non-negative powers of z, a Laurent series includes terms with negative powers. A consequence of this is that a Laurent series may be used in cases where a Taylor expansion is not possible.
2 Calculating the Laurent series expansion
To calculate the Laurent series we use the standard and modified geometric series which are
1 − z
∑^ ∞
n=
zn, |z| < 1 ,
∑^ ∞
n=
zn^
, |z| > 1.
Here f (z) = (^1) −^1 z is analytic everywhere apart from the singularity at z = 1. Above are the expansions for f in the regions inside and outside the circle of radius 1 , centred on z = 0, where |z| < 1 is the region inside the circle and |z| > 1 is the region outside the circle.
|z| < 1
|z| > 1
s^1
2.1 Example Determine the Laurent series for f (z) =
(z + 5) (2)
that are valid in the regions
(i) {z : |z| < 5 }, and (ii) {z : |z| > 5 }.
Solution The region (i) is an open disk inside a circle of radius 5 , centred on z = 0, and the region (ii) is an open annulus outside a circle of radius 5 , centred on z = 0. To make the series expansion easier to calculate we can manipulate our f (z) into a form similar to the series expansion shown in equation (1). So f (z) = 1 5(1 + z 5 )
5(1 − (− z 5 ))
|z| < 5
|z| > 5
− (^5) s
Now using the standard and modified geometric series, equation (1), we can calculate that
f (z) = 1 5(1 − (− z 5 ))
∑^ ∞
n=
− z 5
)n , |z| < 5 ,
∑^ ∞
n=
(^1
z 5
)n , |z| > 5.
Hence, for part (i) the series expansion is
f (z) =^1 5
∑^ ∞
n=
− z 5
)n =^1 5
∑^ ∞
n=
(−1)n^ zn 5 n^
∑^ ∞
n=
(−1)n^ zn 5 n+^
, |z| < 5 ,
which is a Taylor series. And for part (ii) the series expansion is
f (z) = − (^15)
∑^ ∞
n=
(^1
z 5
)n = − 1 5
∑^ ∞
n=
(−1)n^5 n zn^ =^ −
∑^ ∞
n=
(−1)n^5 n−^1 zn^ ,^ |z|^ >^5.
2.2 Example
Determine the Laurent series for f (z) = 1 z(z + 5)
valid in the region {z : |z| < 5 }.
Solution We know from example 2.1 that for
1 (z + 5)
, the series expansion is
∑^ ∞
n=
(−1)n^ zn 5 n+^
, |z| < 5.
It follows from this that we can calculate the series expansion of f (z) as
f (z) =^1 z
(z + 5)
=^1
z
∑^ ∞
n=
(−1)n^ zn 5 n+^
∑^ ∞
n=
(−1)n^ zn−^1 5 n+^
|z| < 5
|z| > 5
− (^5) s s
2.3 Example For the following function f determine the Laurent series that is valid within the stated region R.
f (z) = 1 z(z + 2)
, R = {z : 1 < |z − 1 | < 3 }. (4)
Solution The region R is an open annulus between circles of radius 1 and 3 , centred on z = 1. We want a series expansion about z = 1; to do this we make a substitution w = z− 1 and look for the expansion in w where 1 < |w| < 3. In terms of w
f (z) =
(w + 1)(w + 3).
1 < |z − 1 | < 3
R
− (^2) s (^0) s 1
1 < |w| < 3
− (^3) s − (^1) s
To make the series expansion easier to calculate we can manipulate our f (z) into a form similar to the series expansion shown in equation (1). To do this we will manipulate the fraction into a form based on equation (1). We get
f (z) =
4iw
1 − ( − 4iw )
4iw(1 − iw 4 )
Using the the standard and modified geometric series, equation (1), we can calculate that
4iw(1 − iw 4 )
4iw
∑^ ∞
n=
iw 4
)n
∑^ ∞
n=
(iw)n−^1 4 n+^
, |w| < 4 ,
4iw
∑^ ∞
n=
( iw 4 )n^
4iw
∑^ ∞
n=
iw
)n = −
∑^ ∞
n=
4 n−^1 (iw)n+^
, |w| > 4.
We require the expansion in w where |w| > 4 , so
f (z) = −
∑^ ∞
n=
4 n−^1 (iw)n+1^ =^ −
∑^ ∞
n=
4 n−^2 (iw)n^.
Substituting back in w = z − 2i we get the Laurent series valid within the region |z − 2i| > 4
f (z) = −
∑^ ∞
n=
4 n−^2 (i(z − 2i))n^.
3 Key points
- First check to see if you need to make a substitution for the region you are working with, a substitution is useful if the region is not centred on z = 0.
- Then you will need to manipulate the function into a form where you can use the series expansions shown in example (1): this may involve splitting by partial fractions first.
- Find the series expansions for each of the fractions you have in your function within the specified region, then substitute these back into your function.
- Finally, simplify the function and, if you made a substitution, change it back into the original variable.
For more information on Laurent series refer to the lecture notes.
